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Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

Characteristic equation $ \mid A – \lambda I \mid = 0$

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 1\\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & -\lambda \end{vmatrix} = 0$

Perform the operation, $C_{4} \rightarrow C_{1} + C_{2} + C_{3} + C_{4}$, we get

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 3-\lambda\\ 1& -\lambda & 1 & 3-\lambda\\ 1& 1 & -\lambda & 3-\lambda \\1 & 1 & 1 & 3-\lambda \end{vmatrix} = 0$

$\implies (3 – \lambda) \begin{vmatrix} -\lambda & 1 & 1 & 1 \\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & 1 \end{vmatrix} = 0$

Perform the operation, $R_{1} \rightarrow R_{1} - R_{2} , R_{2} \rightarrow R_{2} - R_{1}, R_{3} \rightarrow R_{3} - R_{1},$ we get

$\implies (3 – \lambda) \begin{vmatrix} -\lambda - 1 & 0 & 0 & 0 \\ 1 + \lambda & -\lambda-1 & 0 & 0 \\ 1 + \lambda & 0 & -\lambda – 1 & 0 \\1 & 1 & 1 & 1 \end{vmatrix} = 0$

$\implies (3-\lambda) (-\lambda – 1)(-\lambda – 1)(-\lambda – 1) = 0$

$\implies \lambda = -1,-1,-1,3$

$\therefore$ The largest eigenvalue is $3.$

$\textbf{PS:}$ For any matrix $A,$

- The determinant of $A$ equals the product of its eigenvalues.
- The trace of $A$ equals the sum of its eigenvalues.
- The trace of a matrix is defined as the sum of the leading diagonal entries.
- A real symmetric matrix has only real eigenvalues.

$$\text(OR)$$

Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

We can write given matrix as $:3\begin{pmatrix} 0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &0 &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &0 &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &0 \end{pmatrix} = 3A$

A Markov matrix is a square matrix with all nonnegative entries, and where the sum of the entries down any column is $1.$ If the entries are all positive, it’s a positive Markov matrix. It is also called a doubly stochastic matrix.

The most important facts about a positive Markov matrix are:

- $\lambda = 1$ is an eigenvalue.
- The eigenvector associated with $\lambda = 1$ can be chosen to be strictly positive.
- All other eigenvalues have a magnitude less than $1.$

So, the correct answer is $3.$

References:

If you understand the structure of the matrix, you can solve it within some seconds.

You can write given matrix as : $3\begin{pmatrix} 0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &0 &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &0 &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &0 \end{pmatrix} = 3A$

This special matrix A is called Doubly Stochastic Matrix and it has largest eigen value $= 1$.

It is also used in Google’s PageRank algorithm.

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